- #1

noboost4you

- 61

- 0

f(x) = (ln x)^x

I know the derivative of ln x is 1/x, but the exponent is throwing me off. Can anyone offering any help? Thanks alot

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- Thread starter noboost4you
- Start date

- #1

noboost4you

- 61

- 0

f(x) = (ln x)^x

I know the derivative of ln x is 1/x, but the exponent is throwing me off. Can anyone offering any help? Thanks alot

- #2

phoenixthoth

- 1,605

- 2

[tex]a^{b}=e^{b\ln a}[/tex]

so [tex]\left( \ln x\right) ^{x}=e^{x\ln \left( \ln x\right) }[/tex]. now use product and chain rules.

so [tex]\left( \ln x\right) ^{x}=e^{x\ln \left( \ln x\right) }[/tex]. now use product and chain rules.

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- #3

Ambitwistor

- 841

- 1

- #4

noboost4you

- 61

- 0

Originally posted by phoenixthoth

[tex]a^{b}=e^{b\ln a}[/tex]

so [tex]\left( \ln x\right) ^{x}=e^{x\ln \left( \ln x\right) }[/tex]. now use product and chain rules.

Would I then have to substitute? I'm still not completely following... Thanks again

Last edited:

- #5

phoenixthoth

- 1,605

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- #6

noboost4you

- 61

- 0

xe^(1/x) ??

i don't know how to use the power and chain rule on e^xln(ln x)

- #7

phoenixthoth

- 1,605

- 2

we have [tex]y=e^{x\ln \left( \ln x\right) }[/tex].

this can be written as [tex]y=e^{u}[/tex] where [tex]u=x\ln \left( \ln x\right) [/tex].

the chain rule is that [tex]\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}[/tex].

[tex]\frac{dy}{du}=e^{u}[/tex]. to find [tex]\frac{du}{dx}[/tex], note that [tex]u[/tex] is the

product of [tex]x[/tex] and [tex]\ln \left( \ln x\right) [/tex].

[tex]\frac{du}{dx}=\left( \frac{d}{dx}x\right) \ln \left( \ln x\right) +x\frac{d

}{dx}\ln \left( \ln x\right) [/tex]. [tex]\frac{d}{dx}x=1[/tex] and to find [tex]\frac{d}{dx}

\ln \left( \ln x\right) [/tex], it may be useful to write [tex]v=\ln w[/tex] where [tex]w=\ln x

[/tex].

[tex]\frac{d}{dx}\ln \left( \ln x\right) =\frac{dv}{dx}=\frac{dv}{dw}\frac{dw}{dx

}.[/tex]

[tex]\frac{dv}{dw}=\frac{1}{w}[/tex] and [tex]\frac{dw}{dx}=\frac{1}{x}[/tex]. hence [tex]\frac{d

}{dx}\ln \left( \ln x\right) =\frac{1}{w}\frac{1}{x}=\frac{1}{\ln x}\frac{1}{

x}=\frac{1}{x\ln x}[/tex].

putting this back into the most recent expression for [tex]\frac{du}{dx}[/tex], we

get [tex]\frac{du}{dx}=1\cdot \ln \left( \ln x\right) +x\left( \frac{1}{x\ln x}

\right) =\ln \left( \ln x\right) +\frac{1}{\ln x}[/tex].

putting this back into the most recent expression for [tex]\frac{dy}{dx}[/tex], we

get [tex]\frac{dy}{dx}=e^{u}\left( \ln \left( \ln x\right) +\frac{1}{\ln x}

\right) =e^{x\ln \left( \ln x\right) }\left( \ln \left( \ln x\right) +\frac{1

}{\ln x}\right) [/tex].

since [tex]\left( \ln x\right) ^{x}=e^{x\ln \left( \ln x\right) }[/tex], we get [tex]

\frac{dy}{dx}=e^{x\ln \left( \ln x\right) }\left( \ln \left( \ln x\right) +

\frac{1}{\ln x}\right) =\left( \ln x\right) ^{x}\left( \ln \left( \ln

x\right) +\frac{1}{\ln x}\right) [/tex]. either the middle or right side of this equation may be acceptable.

this can be written as [tex]y=e^{u}[/tex] where [tex]u=x\ln \left( \ln x\right) [/tex].

the chain rule is that [tex]\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}[/tex].

[tex]\frac{dy}{du}=e^{u}[/tex]. to find [tex]\frac{du}{dx}[/tex], note that [tex]u[/tex] is the

product of [tex]x[/tex] and [tex]\ln \left( \ln x\right) [/tex].

[tex]\frac{du}{dx}=\left( \frac{d}{dx}x\right) \ln \left( \ln x\right) +x\frac{d

}{dx}\ln \left( \ln x\right) [/tex]. [tex]\frac{d}{dx}x=1[/tex] and to find [tex]\frac{d}{dx}

\ln \left( \ln x\right) [/tex], it may be useful to write [tex]v=\ln w[/tex] where [tex]w=\ln x

[/tex].

[tex]\frac{d}{dx}\ln \left( \ln x\right) =\frac{dv}{dx}=\frac{dv}{dw}\frac{dw}{dx

}.[/tex]

[tex]\frac{dv}{dw}=\frac{1}{w}[/tex] and [tex]\frac{dw}{dx}=\frac{1}{x}[/tex]. hence [tex]\frac{d

}{dx}\ln \left( \ln x\right) =\frac{1}{w}\frac{1}{x}=\frac{1}{\ln x}\frac{1}{

x}=\frac{1}{x\ln x}[/tex].

putting this back into the most recent expression for [tex]\frac{du}{dx}[/tex], we

get [tex]\frac{du}{dx}=1\cdot \ln \left( \ln x\right) +x\left( \frac{1}{x\ln x}

\right) =\ln \left( \ln x\right) +\frac{1}{\ln x}[/tex].

putting this back into the most recent expression for [tex]\frac{dy}{dx}[/tex], we

get [tex]\frac{dy}{dx}=e^{u}\left( \ln \left( \ln x\right) +\frac{1}{\ln x}

\right) =e^{x\ln \left( \ln x\right) }\left( \ln \left( \ln x\right) +\frac{1

}{\ln x}\right) [/tex].

since [tex]\left( \ln x\right) ^{x}=e^{x\ln \left( \ln x\right) }[/tex], we get [tex]

\frac{dy}{dx}=e^{x\ln \left( \ln x\right) }\left( \ln \left( \ln x\right) +

\frac{1}{\ln x}\right) =\left( \ln x\right) ^{x}\left( \ln \left( \ln

x\right) +\frac{1}{\ln x}\right) [/tex]. either the middle or right side of this equation may be acceptable.

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- #8

noboost4you

- 61

- 0

Please elaborate. Otherwise, everything else has been very helpful.

- #9

phoenixthoth

- 1,605

- 2

it's based on the property [tex]e^{\ln a}=a[/tex]. if we raise both sides to the [tex]b[/tex] power, we get [tex]\left( e^{\ln a}\right) ^{b}=a^{b}[/tex] which becomes [tex]a^{b}=e^{b\ln a}[/tex]. in this case, [tex]a=\ln x[/tex] and [tex]b=x[/tex].

Last edited:

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